Question

Alguem me poderia ajudar na questao 2 A ?

Answer 1

$Resolu\c{c}\~ao \to \left\{\begin{array}{ccc}log_3\ (\frac{x+3}{x-1}) = 1\\\\3^1 = \frac{x+3}{x-1}\\\\3*(x-1) = x+3\\\\3x-3 = x+3\\\\3x-x = 3+3\\\\2x = 6\\\\\boxed{\boxed{x = \frac{6}{2} = 3}}\end{array}\right$

Espero ter ajudado. =^.^=